When we solve this initial value problem, why for Y is a function of X, we're going to get all the YS on one side, all the XS on the other. Then we're going to use our trigonometric substitution because we can see that this looks like a Pythagorean theorem identity. So the X is going to be on the hypotenuse and the two is on the adjacent side so that we can use secant thetas X / 2 cross multiply we get 2 secant Theta equaling X the derivative each side two secant Theta tangent Theta D Theta equal DX. So when we substitute we get the integral of DY which is Y equaling the integral. Instead of X ^2, we're going to put in four secant squared Theta because if X is 2 secant Theta, then X ^2 is 4 secant squared Theta. We have the -4 that's all going to go over the X which was two secant Theta. We multiply it by the DX which is 2 secant Theta, tangent Theta D Theta. The two secant thetas are going to cancel and the numerator and denominator sqrt 4 is going to give us two. And then we get secant squared Theta -1 left when we had the square root of secant squared Theta -1. That's really just square root of tangent squared Theta, which is tangent Theta. So we end up with the simplifying the two integral tangent squared Theta, D Theta, and we do not know how to take the integral of that. So we're going to rewrite tangent squared Theta as secant squared Theta -1 and we do know how to evaluate each of those terms. The derivative of tangent Theta is secant squared Theta. The derivative of -1 is negative Theta plus RC, so we get 2. The tangent Theta going back to our triangle tangent's going to be the opposite side over adjacent. So if we know two sides of that right triangle, we can find the third side. By our Pythagorean theorem. This is just X ^2 - 4, so our tangent is going to be sqrt X ^2 - 4 / 2 minus Theta. When we look up here, we're going to solve for Theta. So we're going to get Theta being secant inverse of X / 2 + C, multiply through by the two, and then we have an initial value problems. So we're going to put 2IN for the X and four in for the Y. So 4 equals square root 2 ^2 - 4 - 2, secant inverse of 2 / 2 + C This is going to give us 0. The secant inverse of one is 0, so we have C is 4. So we're going to get Y equaling the square root X ^2 - 4 - 2 secant inverse X / 2 + 4.